[R-sig-eco] using Pearson's Chi-squared to verify dependence among species distribuion

[Jari Oksanen]

On 01/05/2013, at 06:09 AM, Antonio Silva wrote:

> Thanks Zoltan R R-sig list members
> 
> I still with some doubts.
> 
> Is there a way to calculate in R the results for the equation
> 
> X2= n*[|ad-bc|-(p/2)]^2 / [(a+b)*(c+d)*(a+c)*(b+d)]
> 
> (see Legendre & Legendre pg 295 eq. 7.6)

This should be doable in vegan::designdist() function. Set 'abcd = TRUE' and you can directly use terms like you defined them above. I won't give the equation here as I don't know what is the equation you used: Your previous message had a different equation than this one here with some mix up with 'n' and 'p'. The designdist() function calculates (dis)similarities between rows. You must transpose (t()) your data if you want to have dissimilarities between columns. The code is pure R so that you can see how to do these calculations yourself.

Cheers, Jari Oksanen

> 
> for the following data:
> 
> 
> ST01 ST02 ST03 ST04 ST05 ST06 ST07 ST08 ST09  SP1 1 1 1 1 1 0 0 1 1  SP2 1 1
> 0 1 0 1 0 1 0
> 
> We have a=4, b=3, c=1, d=1, N=9 and the 2x2 table for this example is
> 
> 
> 
> SP2
> 
> 
> Presence Absent
> SP1 Presence 4 3
> Absent 1 1
> 
> 
> 
> 
> 
> Following the Chi square formula I got 110.25 / 280 = 0.3937
> 
> I have tried many things, without success.
> 
> Thanks in advance for any suggestion.
> 
> Antonio Olinto
> 
> 
> 2013/4/25 Zoltan Botta-Dukat <botta-dukat.zoltan at okologia.mta.hu>
> 
>> Dear Antonio,
>> 
>> Try this:
>> 
>> chisq.test(table(sp1,sp2))
>> 
>> Best wishes
>> 
>> Zoltan
>> 
> 
> 
>>> 
>>> On Wed, Apr 24, 2013 at 6:02 PM, Antonio Silva <aolinto.lst at gmail.com
>>> wrote:
>>> 
>>>> Hi,
>>>> 
>>>> I'm trying to use Pearson's Chi-squared to verify the dependence among
>>>> species distribuion.
>>>> 
>>>> I have a dataframe with the presence/absence data of two species in a
>>>> number of sample units
>>>> 
>>>> The equation I'm using is:
>>>> 
>>>> X2= p*(|ad-bc|-p/2)^2 / ((a+b)*(c+d)*(a+c)*(b+d))
>>>> 
>>>> where a is the number of double presence (1-1), b is the number of 1-0,
>> c
>>>> is the number of 0-1 and d is the number of 0,0 (double absence)
>>>> p is a+b+c+d
>>>> 
>>>> Is there a function to caltulate it using R? I could not understand how
>> to
>>>> use chisq.test function for this.
>>>> 
>>>> Thanks in advance.
>>>> 
>>>> Antonio Olinto
>> 
>> 
>> --
>> 
>> Zoltán BOTTA-Dukát
>> --------------------------------
>> Institute of Ecology and Botany
>> Hungarian Academy of Sciences
>> Centre for Ecological Research
>> --------------------------------
>> H-2163 Vácrátót, Alkomány u. 2-4.
>> HUNGARY
>> Phone: +36 28 360122/157
>> Fax..: +36 28 360110
>> botta-dukat.zoltan at okologia.mta.hu
>> www.okologia.mta.hu
>> 
> 
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-- 
Jari Oksanen, Dept Biology, Univ Oulu, 90014 Finland
jari.oksanen at oulu.fi, Ph. +358 400 408593, http://cc.oulu.fi/~jarioksa